I'm not sure I understand...
I choose n = 5. So, I write 1 up to 10.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
I take any 2: 8 and 9
The absolute difference is 1... which is odd
Write the numbers 1 to 2n, where n is odd, on a white board. You are allowed to perform only the following operation: select 2 numbers, then subtract them and take the absolute value. Is it possible to end up with an odd last number? What about the case where n is even?
Yeah, it's odd. Sorry, my statement of the problem may have been a little bit misleading. My question is: is it ever possible to end with an odd number after eliminating all but 1 of the numbers in this fashion. If it is, is the last number allways going to be odd? Could it possibly be even?
OK; let's have a look at your problem as you posted it...
...and so on
Here's what the puzzle is as you stated:
> Write the numbers 1 to 2n, where n is odd, on a white board. You are allowed to perform only the following operation:
> select 2 numbers, then subtract them and take the absolute value. Is it possible to end up with an odd last number?
n=1: I select 1,2 ; diff = 1
n=2: I select 1,2 then 3,4 then 5,6 ; diff's always 1
n=3: I select 1,2 then 3,4 then 5,6 then 7,8 then 9,10 ; diff's = always 1
So I end up with an odd number in ALL cases.
I don't know what else to tell you.
Okay. I apologize. I should have reread my initial post after your initial reply. I forgot to add a critical part of the problem to my post. After you perform the operation, you erase the 2 numbers that you used with the result of the operation. You continue this process until you have one remaining number. Will the process always terminate with an odd number? An even number? Is it possible for it to terminate with an odd number? An even number?
Ya UnKnown; stuff like:
select 6,3: 1,2,3,4,5
select 5,2: 1,3,3,4
select 1,4: 3,3,3
select 3,3: 0,3 ??????
Wonder if he means:
1: numbers from 1 to 2n where n is an odd integer are written down
2: you select any 2 numbers you wish and let k = absolute difference of these 2 numbers
3: you then remove the 2 numbers from the list, and insert k
4: you continue doing this until the last number in the list is the last inserted k
5: can you devise a strategy where k will always be odd?
(would need a rule applying to zero results)
Considering only the even/odd parity of the numbers (i.e. numbers modulo 2), it does not matter whether we are adding or subtraction, or taking absolute value. The last number is some arithmetic addition of all 2n numbers where some are positive and some negative.
The ending parity will be the same as the parity of the sum of numbers 1 to 2n.
Which is (2n + 1)n, and this is always odd for odd n.